# Thread: Find all y in [0,2π]

1. ## Find all y in [0,2π]

Find all y in [0,2π]

a. $\displaystyle 2(cosx)^2-\sqrt{3}\left(\frac{1}{secx}\right) = 0$
$\displaystyle 2cos^2 - \sqrt{3}\left(\frac{\frac{1}{1}{cosx}}\right) = 0$
$\displaystyle 2cos^2 - \sqrt{3} cos x= 0$
How do I progress from here?

b. $\displaystyle cos^2x + cosx = sin^2x$
$\displaystyle cos^2x+cosx-sin^2x=0$

$\displaystyle 2cos^2x+cos=0$

$\displaystyle cosx(2cosx+1)=0$

$\displaystyle cosx=0$

$\displaystyle 2cosx+1=0$

$\displaystyle cosx=\frac{-1}{2}$

$\displaystyle y = \frac\pi2;\frac{3\pi}{2}; \frac{2\pi}{3}; \frac{4\pi}{3}$

3. ## Re: Find all y in [0,2π]

Cos^2x + cos x = sin ^2 x
= 1 – cos^2 x
That gives
2 cos ^2 x + cos x -1 = 0
(Cos x + 1)( 2cosx – 1 )= 0
Cos x = -1 OR cos x = ½
Now you can proceed further

4. ## Re: Find all y in [0,2π]

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